What is 4x^2+y^2<= 17 ?

The solution to 4x^2+y^2<= 17 is Ellipse with (h,k)=(0,0),a=(sqrt(17))/2 ,b=sqrt(17)

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4x^2+y^2<= 17

4 x^{2} + y^{2} \leq 17

(h, k) = (0, 0), a = \frac{17}{2}, b = 17

Solution steps

4 x^{2} + y^{2} \leq 17

Rewrite

4 x^{2} + y^{2} \leq 17

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{17}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 17 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{17}{2}, b = 17

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 17, a = \frac{17}{2}

What is 4x^2+y^2<= 17 ?
The solution to 4x^2+y^2<= 17 is Ellipse with (h,k)=(0,0),a=(sqrt(17))/2 ,b=sqrt(17)